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3 years ago

What are real images? Are the upright or inverted?

Physics

1 answer:

Mars2501 [29]3 years ago

6 0

Answer:

A real image can be viewed when a screen is placed in the plane of convergence. Real images are inverted... HOPE THIS HELPS!!!

Explanation:

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Assume that when you stretch your torso vertically as much as you can, your center of mass is 1.0 m above the floor. The maximum

Elenna [48]

1) 0.77 m

2) 0.23 m

Explanation:

Here we want to find the time elapsed for crouching in order to jump and reach a height of 2.0 m above the floor, starting from 1.0 m above the floor.

First of all, we start by calculating the speed required to jump up to a height of 2.0 m. Since the total energy is conserved, the initial kinetic energy is converted into gravitational potential energy, so:

$\frac{1}{2}mv^2 = mgh$

where

m is the mass of the man

v is the speed after jumping

$g=9.8 m/s^2$ is the acceleration due to gravity

h = 2.0 - 1.0 = 1.0 m is the change in height

Solving for v,

$v=\sqrt{2gh}=\sqrt{2(9.8)(1.0)}=4.43 m/s$

In the acceleration phase, we know that the initial velocity is

$u=0$

And the force exerted on the floor is 2.3 times the gravitational force, so

$F=2.3 mg$

This means the net force on you is

$F_{net} = F-mg=2.3mg-mg=1.3 mg$

because we have to consider the force of gravity acting downward.

So the acceleration of the man is

$a=\frac{F_{net}}{m}=\frac{1.3mg}{m}=1.3g$

Now we can use the following suvat equation to find the displacement in the acceleration phase, which is how low the man has to crouch in order to jump:

$v^2-u^2=2as$

where s is the quantity we want to find. Solving for s,

$s=\frac{v^2-u^2}{2a}=\frac{4.43^2-0}{2(1.3g)}=0.77 m$

At the beginning, we are told that the height of the center of mass above the floor is

h = 1.0 m

During the acceleration phase and the crouch, the height of the center of mass of the body decreases by

$\Delta h = -0.77 m$

This means that the lowest point reached by the center of mass above the floor during the crouch is

$h'=h+\Delta h = 1.0 - 0.77 = 0.23 m$

This value seems unpractical, since it is not really easy to crouch until having the center of mass 0.23 m above the ground.

3 0

3 years ago

Which of the following has three electrons in its valence shell? carbon aluminum sodium calcium

JulijaS [17]

Aluminum has 3 valence electrons.
<span>You can figure out the valence electrons by looking at its group number. Aluminum is in group 3, so it has 3 valence electrons. </span>

6 0

3 years ago

Read 2 more answers

What is the rotational inertia of an object that rolls without slipping down a 2.00-m-high incline starting from rest, and has a

krek1111 [17]

We will apply the concepts of energy conservation to solve the problem. We know that gravitational potential energy is equivalent to the sum of translational kinetic energy and rotational kinetic energy. Additionally, the relation of the angular velocity with the tangential velocity will be determined to eliminate the angular term and obtain the expression of the Inertia in terms of the data given, therefore, we know that

$PE_{g} = KE_{trans} +KE_{rot}$

$mgh = \frac{1}{2} mv^2+\frac{1}{2}I\omega^2$

The angular velocity in terms of tangential velocity and radius is defined as,

$\omega = \frac{v}{R}$

Replacing,

$mgh = \frac{1}{2} mv^2+\frac{1}{2} I(\frac{v}{R})^2$

Multiplying by R^2,

$gh(mR^2) = \frac{v^2}{2} (mR^2)+\frac{v^2}{2}I$

$\frac{v^2}{2}I = (gh-\frac{v^2}{2})mR^2$

$I = (\frac{2gh}{v^2}-1)mR^2$

Replacing with our values we have,

$I = (\frac{2(9.8)(2)}{6^2}-1)mR^2$

$I = 0.089mR^2$

6 0

3 years ago

A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts "Help.

uranmaximum [27]

Answer:

532.0725 m

102.17270893 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² = g

H = Height of cliff

Distance traveled in 3 seconds

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 3^2\\\Rightarrow s=44.145\ m$

Distance traveled by sound = 2H-44.145 m

$2H-44.145=ut+\dfrac{1}{2}at^2\\\Rightarrow 2H-44.145=340\times 3\\\Rightarrow H=\dfrac{340\times 3+44.145}{2}\\\Rightarrow H=532.0725\ m$

The height of the cliff is 532.0725 m

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 532.0725+0^2}\\\Rightarrow v=102.17270893\ m/s$

Her speed just before she hits the ground is 102.17270893 m/s

5 0

3 years ago

A crate of books is to be put on a truck with the help of some planks sloping up at 36◦ . The mass of the crate is 80 kg, and th

Dafna11 [192]

Answer:

F= 2.86KN

Explanation:

The attached diagram shows the motion of crate on a single line The forces acting on it have been resolved as shown in the attached figure.The kinetic friction opposes the motion.Now crate is moving with constant speed so the acceleration is constant.

Applying Newton's second law to determine the force F and also substituting the values for $F_{k}$ to eliminate normal force